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Spin and spatial exchange symmetry. Write the normalized spin singlet and triplet states asThe singlet is antisymmetric under exchange and has total spin zero; the triplet is symmetric and has total spin one. For orthogonal orbitals defineIdentical spin one-half particles are fermions, so the complete state must be antisymmetric. Thus $C{\chi }_{00}$ is allowed, and $A{\chi }_{1M}$ is allowed for each triplet component. The spatially symmetric same-orbital states $|aa⟩,|bb⟩$ also require the singlet. A same-orbital triplet vanishes by the Pauli exclusion principle.

These products give a spin-adapted basis. More generally the physical state can be writtenA superposition of the singlet and triplet sectors need not factor into one spatial function times one spin function. Factorization holds for a definite-spin sector, as used in the subsequent model. If the original normalized orbitals are not orthogonal, the corresponding $ab\pm ba$ normalization is $\sqrt{2(1\pm |⟨a|b⟩{|}^2)}$; the remainder of this calculation uses the stipulated orthogonal orbitals.

Integral matrix elements and positivity. Let $a(\mathbf{r})$ and $b(\mathbf{r})$ denote the localized spatial orbitals. With fixed proton positions, the one-electron electronic operator isThe proton-proton repulsion is a separate common nuclear energy. The requested one-electron and direct two-electron elements areSince $H_1=h\otimes I$,Likewise,The integrand is nonnegative and is positive on a set of nonzero measure. This is the repulsion cost of double occupancy; the coincidence singularity is integrable for ordinary smooth localized orbitals.

One electron on each center. Suppress the normalized singlet spin factor and use $|\psi ⟩=C$. For each of the spatial products $ab$ and $ba$, the one-body sum has expectation $2E_0$. The cross terms vanish because a one-body hop leaves an orthogonal spectator orbital. The interaction contribution is the direct integral plus the exchange integral, both neglected in the model. ConsequentlyThis state has no same-center double occupancy, but remains an exchange-symmetric spatial superposition rather than assigning a distinguishable electron permanently to each proton.

Two electrons in the bonding orbital. The normalized one-electron bonding orbital is $g=(a+b)/\sqrt{2}$. The trial spatial state is $|\varphi ⟩=|gg⟩$, whose one-body expectation isExpanding the product gives $|\varphi ⟩=(|aa⟩+|ab⟩+|ba⟩+|bb⟩)/2$. The total probability of double occupancy is therefore $1/2$. Keeping only on-site Coulomb repulsion gives the intended interaction truncation,and the interaction expectation is $U/2$. HenceThis truncation must be stated. In particular, mixed Coulomb elements such as $X=⟨aa|H_{1,2}|ab⟩$ do not vanish merely by orbital orthogonality or inversion symmetry. The printed list does not explicitly name them. If they are retained and are equal and real by site symmetry, their contribution to $E_{\varphi }$ is $2X$; the requested expression uses their neglect along with the other off-site interaction terms. No numerical value of $X$ is supplied or invented.

Energy comparison and the bond. Put $x=|t|/U$ with $U>0$ and $t=-|t|$. ThenThe straight line starts at $1/2$ and crosses zero at $x=1/4$. The purely bonding-orbital trial state is better for $x>1/4$, while the singly occupied covalent trial state is better for $x<1/4$.

For localized orbitals of approximately fixed atomic size, the hopping magnitude $|t|$ decreases rapidly, typically exponentially times a polynomial, as the proton separation grows. The on-site integral $U$ remains of atomic Coulomb scale and tends to the isolated-orbital value; it does not decay as the intersite repulsion $1/R$. If the orbitals themselves are allowed to relax, $U$ also changes, so no universal monotonic law for $U(R)$ follows from the data.

At small $|t|/U$, electron correlation suppresses the ionic configurations $aa$ and $bb$, favoring a singlet with roughly one electron on each center. A single doubly occupied bonding orbital has an artificial finite ionic weight even as the centers separate. At large $|t|/U$, delocalization gains enough kinetic energy that the bonding-orbital description becomes useful and charge fluctuations are substantial.

The crossing of two restricted trial energies is not a sharp transition of the actual two-site ground state. To see this without adding empirical data, include both $C$ and the symmetric ionic state $I_{+}=(|aa⟩+|bb⟩)/\sqrt{2}$. This two-site Hubbard dimer reduces toIts ground state is $\cos \theta C+\sin \theta I_{+}$, with $\tan 2\theta =4|t|/U$ and $0\le \theta \le \pi /4$. The ionic probability isIt changes smoothly from zero to $1/2$. At large separation, the electronic lowering is $-4t^2/U$ to leading order and vanishes as hopping vanishes. An absolute molecular binding curve would additionally require $E_0(R)$, the nuclear repulsion and the neglected intersite terms; this simplified comparison alone does not determine an equilibrium bond length.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

A coherent state is a normalized eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding in the Fock state eigenbasis gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and wavefunction normalization fixesThe displacement operator therefore shifts the oscillator ground state in phase space. Distinct coherent states are not orthogonal:

The occupation probabilities form a Poisson distribution,For the quantum harmonic oscillator, the expectation values of position and momentum areThe fluctuations equal those of the ground state,Thus the position wavefunction is a displaced Gaussian wave packet that saturates the Heisenberg uncertainty principle.

Time evolution under $H=\hslash \omega (a^{†}a+1/2)$ preserves this form:The center follows the classical oscillator trajectory while the packet width stays constant. Its mean energy is $\hslash \omega (N+1/2)$; it is generally a superposition of energy eigenstates, rather than a state of definite energy. For $N\gg 1$, the relative occupation fluctuation is $1/\sqrt{N}$, explaining the classical appearance of a highly excited coherent state.

For the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

For physical spin ladder operators, $S_j^{\pm }=S_j^x\pm i{S}_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and givesAdding the remaining component provesFor spin one-half, $S^{+}|\downarrow ⟩=\hslash |\uparrow ⟩$, $S^{-}|\uparrow ⟩=\hslash |\downarrow ⟩$, and raising an up spin or lowering a down spin gives zero. These formulas restore the physical $\hslash$ factors implicit in the dimensionless ladder hint.

Let $|n⟩$ abbreviate the state with the only down spin at site $n$, keeping $|{\psi }_0⟩$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up spins. The ladder terms vanish, while $S_j^z{S}_{j+1}^z$ gives ${\hslash }^2/4$. ThusFor a down spin at $j$, the $z$ product is instead $-{\hslash }^2/4$, and $S_j^{+}{S}_{j+1}^{-}$ moves it to $j+1$ with amplitude ${\hslash }^2$. For a down spin at $j+1$, the other ladder product moves it back. ThereforeEvery bond in $|{\psi }_0⟩$ is aligned, so summing its $N$ bond energies givesThis is indeed a ground-state energy: a pair of spin-one-half particles has ${\mathbf{S}}_j\cdot {\mathbf{S}}_{j+1}$ eigenvalues ${\hslash }^2/4$ and $-3{\hslash }^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic Heisenberg model has a whole maximum-total-spin ground multiplet, not a unique all-up ground state.

In the single-excitation subspace, only the two bonds adjacent to the down spin differ from the aligned background. Their diagonal energy increase is $J$, and each neighboring hop has amplitude $-J/2$. Hence, with periodic site indices,This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n⟩=|n+1⟩$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, soThere are $N$ independent such wavevectors, for example one set of $N$ consecutive integers $\ell$ in a Brillouin zone. The normalized one-magnon Bloch states on a ring are $|q⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{iqna}|n⟩$. Reindexing with the periodic coefficient condition gives $T|q⟩=e^{-iqa}|q⟩$ and hence $T^N|q⟩=|q⟩$.

There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation eigenstate, equivalently a consistently periodic plane wave. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed momentum modes.

Apply the hopping Hamiltonian operator and shift the summation index using periodic boundary condition. The coefficients from $|n-1⟩$ and $|n+1⟩$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thusand the nearest-neighbour ferromagnetic magnon dispersion isThe unnormalized state in the question has the same energy. Distinct allowed momenta are orthogonal, since $N^{-1}{\sum }_n{e}^{i(q-q^{\prime })na}={\delta }_{q{q}^{\prime }}$ over one set of representatives.

For large $N$ the sampled wavevectors become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos (qa)$ over $-2\pi <qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm \pi /a$. Near any equivalent zero,using $q$ measured from that zero. The low-momentum ferromagnetic magnon is quadratic, with group velocity ${\hslash }^{-1}dE/dq=(Ja/\hslash )\sin (qa)$. The $q=0$ state is proportional to the total spin lowering operator acting on $|{\psi }_0⟩$, so it belongs to the same ground multiplet and has exactly zero excitation energy. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.

For a self-adjoint Hamiltonian operator bounded below with ground-state energy $E_0$, every admissible nonzero trial wavefunction has a Rayleigh quotient at least $E_0$. To see this explicitly in a complete orthonormal eigenbasis, write $|\psi ⟩={\sum }_n{c}_n|n⟩$ with $H|n⟩=E_n|n⟩$. ThensoFor continuous spectral components, the spectral theorem replaces the sum by an integral with the same nonnegative integrand. Equality requires the trial quantum state to lie entirely in the ground-state subspace. The variational method chooses a tractable family, evaluates the quotient and minimizes over its parameters. This gives an upper bound; enlarging the family can only improve that bound. The trial must belong to the operator domain for the operator expression above, or have finite energy in the corresponding quadratic-form version of the quantum variational principle.

For the oscillator, use the standard kinetic operator $T=-{\hslash }^2{d}^2/(2md{x}^2)$: the original PDF omits the superscript $2$ on the derivative in its numerator. Take $a>0$, and normalize the scaled exact ground state asThe factor $\sqrt{\lambda }$ is optional in a quotient but makes the calculation transparent. Changing variable to $u=\lambda x$ and differentiating the scaled wavefunction givesAt $\lambda =1$ this trial is the exact minimizing state, so its energy $\mathcal{E}(\lambda )={\lambda }^2{T}_0+{\lambda }^{-2}{V}_0$ has zero derivative there. Hence $2T_0-2V_0=0$. Since $E_0=T_0+V_0$,This is the quantum virial identity from scaling for a quadratic potential, and hence the oscillator form of the virial theorem.

For the inverse-power attraction, take $a>0$ and suppose, for contradiction, that an exact ground state exists with separately finite kinetic energy $T_0>0$ and potential mean $U_0<0$, admitting the same dilation variations. Homogeneity now givesStationarity at $\lambda =1$ would require $2T_0+p{U}_0=0$. Its second derivative would then beFor $p>2$ this is negative, contradicting a ground-state minimum. More strongly, one does not need a presumed exact state: choose any normalized smooth test function supported away from the singular origin. Its $T$ and $U$ are finite, and its scaled quotient satisfiesThe unregularized attraction therefore exhibits fall to the centre for a supercritical inverse-power potential and has no stable ground state.

There is a necessary qualification to the PDF's broader wording about “no bound state”. Under the usual finite-energy virial theorem assumptions, a putative negative-energy bound state would satisfywhich contradicts being below the continuum threshold at zero. Thus no negative-energy bound state satisfying those regularity assumptions is possible. But this singular differential expression needs a specified domain at $x=0$; separately divergent kinetic and potential means or boundary conditions not preserved by dilation invalidate that ordinary calculation. The domain assumptions in singular-potential virial arguments must therefore be stated: the scaling proof establishes absence of a semibounded ground-state problem, rather than a domain-independent theorem about every formal square-integrable solution. A short-distance regularization would be a different potential.

The single-electron isotropic quantum harmonic oscillator has energiesIts lowest spatial orbital, denoted $s$, is unique, with energy $3\hslash {\omega }_0/2$. Including spin one-half gives two spin orbitals. Two identical electrons can occupy both only with opposite spin in the unique singlet state. Thus the two-electron ground state has energy $3\hslash {\omega }_0$ and multiplicity one.

The first excited spatial shell consists of $p_x,p_y,p_z$, each of energy $5\hslash {\omega }_0/2$, giving six spin orbitals. A first-excited two-electron Slater determinant occupies one of the two ground-shell spin orbitals and one of these six excited-shell spin orbitals. HenceNo further factor of two is inserted for interchanging the electrons, since the fermionic exchange symmetry is already built into a Slater determinant.

The spatial Hamiltonian operator is invariant under simultaneous rotations of both positions, so $[H_0,L_k]=0$ for each component of the total orbital angular momentum. It is also spin independent, so $[H_0,S_k]=0$ for each total spin component. Thereforeexpressing conservation of angular momentum and conservation of the total spin magnitude for this model. The Pauli exclusion principle restricts which states occur but does not break either symmetry.

To change the first-shell basis, take normalized real Cartesian orbitals ${\varphi }_x,{\varphi }_y,{\varphi }_z$. Using spherical coordinates, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$. Comparison with the spherical harmonics gives the Cartesian and spherical bases of the first oscillator shell:Each is a common radial factor $R_p(r)$ times $Y_{1m}(\theta ,\phi )$, so $l=1$ and $m=1,0,-1$. Their normalizations and mutual orthogonality follow because this is a unitary change of basis. The ground spatial orbital has $l=0$.

For two ground-shell electrons, addition of angular momentum gives $L=0$, and their symmetric spatial state requires $S=0$. For one electron in each shell, $0\otimes 1$ gives only $L=1$. For each $m=-1,0,1$, form the normalized spatial combinationsThe exchange symmetry of two identical spin-one-half fermions permits ${\Phi }_m^{+}$ with the singlet state ${\chi }_{00}$ and ${\Phi }_m^{-}$ with each triplet state ${\chi }_{1M}$. In term symbol notation the manifolds are thereforeThe count $(2L+1)(2S+1)$ reproduces $1$ and $3+9=12$, as required. Both exchange symmetries are possible for the distinct occupied orbitals; dropping either spin sector would undercount the two-electron states in the first two shells of an isotropic oscillator.

For the specified spin-orbit interaction, $\mathbf{S}$ and $\mathbf{L}$ are the total two-electron operators. Set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity givesFor the $L=1,S=0$ singlet, $J=1$ and the shift is zero. For the $L=S=1$ triplet, $J=0,1,2$. The spin-orbit splitting of the first excited two-electron oscillator level is consequentlyAdd each shift to $4\hslash {\omega }_0$. The multiplicities sum to twelve and their weighted shift sums to zero. For positive $\lambda$ the table is ordered in energy; for negative $\lambda$ the ordering reverses. Weak coupling means $|\lambda |{\hslash }^2\ll \hslash {\omega }_0$. Since this total-spin interaction commutes with $H_0$, $L^2$ and $S^2$, the shifts within this invariant manifold are in fact exact for the specified model, although only first order was requested.

A confined ground state requires $k>0$; take $\alpha >0$ so the trial wavefunction is normalizable. Its normalized form isThe Gaussian moment recurrence gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Also, integration by parts gives $⟨p^2⟩={\hslash }^2\int |f_{\alpha }^{\prime }{|}^{2}dx={\hslash }^2\alpha /2$. The Rayleigh quotient for the quartic oscillator is consequentlyDifferentiation yields ${\mathcal{E}}^{\prime }(\alpha )={\hslash }^2/(4m)-3k/(2{\alpha }^3)$, so the unique minimum is at ${\alpha }_{\ast }=(6mk/{\hslash }^2{)}^{1/3}$. Since the quotient diverges at both endpoints $0$ and $\infty$, this is the global minimum in the trial family. The Rayleigh-Ritz variational principle givesFor the first excited energy eigenstate, choose the odd trial familyThe even confining potential has an even, nodeless ground state, by the nodeless theorem for a one-dimensional ground state. By the Sturm oscillation theorem, its first excited state is odd with one node. The proposed family has the same parity operator eigenvalue and one node, and is orthogonal to the exact ground state, not merely to its Gaussian trial approximation. Minimization in the odd linear subspace therefore bounds the first excited energy from above. For completeness, its Rayleigh quotient is $3{\hslash }^2\beta /(4m)+15k/(4{\beta }^2)$, minimized at ${\beta }^3=10mk/{\hslash }^2$, giving $E_1\le (9{\hslash }^2/8m)(10mk/{\hslash }^2{)}^{1/3}$. These are Gaussian variational bounds for a quartic oscillator.

If $k=0$, there is no normalizable free particle ground state on the whole line, though the spectral infimum is zero. If $k<0$, the Hamiltonian operator is not bounded below, so a ground-energy bound of the requested kind does not exist.

For the requested oscillator response, treat $y$ as the fixed transverse position of a one-dimensional quantum harmonic oscillator along $z$, and put $\theta =ky$. The supplied radiation is transverse and already satisfies the Coulomb gauge. Its perturbation isThe annihilation operator and creation operator giveThus the only first-order transition out of the ground state is to $|1⟩$. More generally, the oscillator electric-dipole selection rule is $\Delta n=\pm 1$. This is a rule for matrix elements; exact energy conservation is replaced by a finite-time resonance profile during a switched finite-duration interaction.

To display both the resonant and nonresonant terms without spurious singularities, defineThe interaction picture coefficient in the given expansion is thenThe full Schrödinger picture coefficient of $|1⟩$ is $e^{-3i{\omega }_{0}t/2}{b}_1(t)$, while that of $|0⟩$ is $e^{-i{\omega }_{0}t/2}$ at this order. The normalization changes only at order $A_0^2$. In particular, at $\omega ={\omega }_0$ the first term is $C{e}^{i\theta }t$, finite for every finite time, and the perturbative result requires $|b_1|\ll 1$.

The fixed-$y$ interpretation is needed for the displayed target formulae. If $p^2$ were instead a three-dimensional kinetic operator with no transverse confinement, that model would have no normalizable transverse ground state, and $e^{\pm iky}$ would transfer transverse momentum $\pm \hslash k$. A treatment of that different model needs specified transverse states and recoil energies; they are absent from the question. The following parts use the intended localized-oscillator response.

Use zero-based labels to agree with ${\varphi }_0$ in the later parts. The normalized infinite square well energy eigenstates and energy eigenvalues areThe position matrix elements in a symmetric infinite square well, obtained by integrating the product of the two wavefunctions against $x$, areThe integral in the question uses positive integer labels; replacing them by $k+1,n+1$ is essential here. Only opposite-parity states, $k+n$ odd, contribute, as also follows from the parity operator selection rule.

Second-order nondegenerate perturbation theory now gives the requested infinite sum:In fully explicit form this isThe symbol $m$ here denotes the electron mass, not a summation index. For the ground state $n=0$, every denominator in the first form is negative, so the ground energy level moves downward. For excited states, contributions from lower and higher energy levels have opposite signs.

Interpret the two displayed exchange components as belonging to the common nearest-neighbour sum with coefficient $-J$. Write $H_f=|h|$ and, for $h\ne 0$, $s=\mathrm{sgn}h$. The source explicitly fixes $J>0$ but does not separately constrain $h$ in this question; keeping $H_f$ covers either field direction.

For classical vectors of length $S$, define ${\mathbf{q}}_j=(S_j^x,S_j^y)$. The Cauchy-Schwarz inequality gives ${\mathbf{q}}_j\cdot {\mathbf{q}}_{j+1}\le |{\mathbf{q}}_j||{\mathbf{q}}_{j+1}|\le S^2$, and $h{S}_j^x\le H_{f}S$. Each term is therefore bounded below, soUniform alignment along the field saturates every bound:Equality requires all planar vectors to be parallel with length $S$, and the nonzero field then fixes their direction. At $h=0$, every uniformly oriented vector in the $xy$ plane is a classical ground state:A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.